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Since $S^n$ is simply connected for $n \geq 2$ and the identity is a covering map, the identity will do. But note that the universal cover is not uniquely determined (it is only unique up to an isomorphism of covering spaces), so there is no the map. In fact, any self-homeomorphism $\phi$ of $S^{n}$ will be a universal cover $\phi: S^{n} \to S^n$.

The case $n = 1$ is different, since $S^1$ is not simply connected. Its universal cover is e.g. given the exponential map $t \mapsto e^{2\pi i t}$ from $\mathbb{R} \to S^1 = \{z \in \mathbb{C} \,:\,|z| = 1\}$. For every $k \in \mathbb{Z} \smallsetminus \{0\}$ the map $z \mapsto z^{k}$ will determine a self-covering $S^1 \to S^1$ but none of these can be a universal covering.